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尝试从SQL返回true或false

[英]Trying to return true or false from SQL

Here is my code: 这是我的代码:

$result_username = mysqli_query($dbconnection, Data::checkForUsername($username));

It then goes here: 然后转到这里:

   public static function checkForUsername($username) {
            global $database;

            $query = "
                SELECT COUNT(*) 
                FROM users 
                WHERE username='{$username}';";

            $result = $database -> query($query);

            return $result;
        }

Then $result does this: 然后$ result这样做:

if (mysqli_result($result_username, 0) > 0) {

However, it then gives me back a Resource_Id?? 但是,它然后给了我一个Resource_Id? I can not figure out why? 我不知道为什么?

I simply want to check if the username exists in at least 1 row. 我只想检查用户名是否至少存在1行。

You need to fetch your data after you execute your query 执行查询后,您需要获取数据

$row = $result->fetch_array(MYSQLI_NUM);
return $row[0];

UPDATE: Now, using prepared statement your function can look like this: 更新:现在,使用准备好的语句,您的函数可以如下所示:

public static function checkForUsername($username) {
    global $database;

    $result = 0;
    $query = "SELECT COUNT(*) FROM users WHERE username=?";
    /* create a prepared statement */
    if ($stmt = $database->prepare($query)) {
        /* bind parameters for markers */
        $stmt->bind_param("s", $username);
        /* execute query */
        $stmt->execute();
        /* bind result variable */
        $stmt->bind_result($result);
        /* fetch value */
        $stmt->fetch();
        /* close statement */
        $stmt->close();
    }
    return $result;
}

You can try 你可以试试

return $result->num_rows

Then 然后

if (checkForUsername('Redeyes') != 0) {
}

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